INORGANIC CHEMISTRY
1. CHEMICAL CALCULATION
Formula Weight (FW) or Formula Mass
The formula weight of a substance is the sum of the atomic weights of all atoms in a formula unit of the compound, whether molecular or not.
Sodium chloride (NaCl), has a formula weight of 58.44 amu (22.99 amu from Na plus 35.45 amu from Cl). NaCl is ionic, so strictly speaking the expression "molecular weight of NaCl" has no meaning. On the other hand, the molecular weight and the formula weight calculated from the molecular formula of a substance are identical.
Solved Problem
Calculate the formula weight of each of the following to three significant figures, using a table of atomic weight (AW):
(b) Iron
(c) iron sulfate - Fe2 (SO4)3.Solution
(a) chloroform- CHCl3
1 x AW of C = 12.00 amu
1 x AW of H = 1.00 amu
3 x AW of Cl3 = 3 x 35.45 = 106.35 amu
(b) Iron (Fe)
Shown here the atomic numbers (AW) of Carbon and Hydrogen atom |
(c) Iron sulfate (Fe2 (SO4)3
2 x AW of Fe2 = 2 x 55.85 = 111.70 amu
1 x AW of S = 32.07 amu
4 x AW of O = 4 x 16 = 64 amu
i.e. 3 x AW of SO4 = 3 x 96.07 amu
= 288.21 amu
Formula weight of Fe2 (SO4)3 = 111.70 + 288.21 = 399.91 amu
The answer rounded to three significant figures is 4.00 x 102 amu
Calculate the formula weights of the following compounds
(a) NO2
1 x AW of N = 14.01 amu
2 x AW of O = 2 x 16 = 32 amu
Formula weight of NO2 = 14.01 + 32.00 = 46.01 amu
(b). glucose (C6H12O6)
6 x AW of C = 6 x 12.00 = 72 amu
12 x AW of H = 12 x 1.008 = 12.10 amu
6 x AW of O = 6 x 16 = 96 amu
Formula weight of C6H12O6 = 72 + 12.10 + 96 = 180.10 amu
(b) NaOH
1 x AW of Na = 22.99 amu
1 x AW of O = 16.00 amu
1 x AW of H = 01.008 amu
Formula weight of NaOH = 22.99 + 16.00 + 01.008 = 40 amu
(c) Mg(OH)2
1 x AW of Mg = 24.31 amu
2 x AW of OH = 2 (16 + 1.008) = 34.016 amu
Formula weight of Mg(OH)2 = 24.31 + 34.016 = 58.33 amu
(d) Methanol (CH3 OH)
1 x AW of C = 12.01 amu
3 x AW of H = 3.02 amu
1 x AW of O = 16 amu
1 x AW of H = 1.008 amu
Formula weight of NaOH = 12.01 + 3.02 + 16 + 1.008 = 32.04 amu
(e) PCl3
1 x AW of P = 30.97 amu
3 x AW of Cl = 3 x 35.45 = 106.35 amu
Formula weight of Mg(OH)2 = 30.97 + 106.35 = 137.32 amu
(f) K2 CO3
2 x AW of K = 2 x 39.10 = 78.2amu
1 x AW of C = 12.01 amu
3 x AW of O = 3 x 16 = 48 amu
Formula weight of K2 CO3 =78.2 + 12.01 + 48 = 138.21 amu
Avogadro's Number (NA)
The number of atoms in a 12-g sample of carbon - 12 is called Avogadro's number (to which we give the symbol NA).
Recent measurements of this number give the value 6.0221367 x 1023, which is 6.023 x 1023.
A mole of a substance contains Avogadro's number of molecules.
For example a dozen eggs equals 12 eggs, a gross of pencils equals 144 pencils and same like a mole of ethanol equals 6.023 x 1023 ethanol molecules, a mole of methanol equals 6.023 x 1023 methanol molecules
Significance
The molecular mass of SO2 is
Molicular mass of S = 32 g mol-1
Molicular mass of O2 = 32 g mol-1
Ie 64 g mol-1.
64 g of SO2 contains 6.023 x 1023 molecules of SO2.
2.24 x 102m3 of SO2 at S.T.P. contains 6.023 x 1023 molecules of SO2.
Similarly the molecular mass of CO2 is 44 g mol-1. 44g of CO2 contains 6.023 x 1023 molecules of CO2. 2.24 x 10-2m3 of CO2 at S.T.P contains 6.023 x 1023 molecules of CO2.Avogadro's number = 6.0221415 × 1023
Mole concept
One of the values of the mole is that every mole of any element has the same number of atoms in it. We have not yet discussed what that number is in this course (we will deal with that later when we consider the size of atoms), but we do know that it is the same amount for each element.
For example, if you take the equal number of oxygen (Weight of O is 16 amu) and hydrogen (weight of H is 1amu) atom, the weight of hydrogen atom would come 16 times greater of oxygen atom. That means weight of 1 atom of oxygen = 16 x weight of hydrogen atom
Conversely, 16 times more oxygen (by weight) than hydrogen, you will have equal amounts (by atoms) of each. Thus in this case, if we weigh out one gram of hydrogen atom, have 1 mole of hydrogen atom. As same way if we weigh out 16 gram of oxygen atom, have 1 mole of oxygen atom
Consider the reaction: 2 H2 + O2 2H2O
In this reaction one molecule of oxygen reacts with two molecules of hydrogen. So it would be desirable to take the molecules of H2 and oxygen in the ratio 2:1, so that the reactants are completely consumed during the reaction. But atoms and molecules are so small in size that is not possible to count them individually.
In order to overcome these difficulties, the concept of mole was introduced. According to this concept number of particles of the substance is related to the mass of the substance.The mole is defined as the amount of substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as there are atoms in 12 g of the carbon (12-isotope). Thus, by definition, one mole of C has a mass of exactly 12 g
i.e, one mole of an atom consists of Avogadro number of particles
One mole = 6.023 x 1023 particles
One mole of oxygen molecule = 6.023 x 1023 oxygen molecules
One mole of ethanol molecule = 6.023 x 1023 ethanol molecules
In using the term mole for ionic substances, we mean the number of formula units of the substance.
For example, a mole of sodium carbonate, Na2CO3 is a quantity containing 6.023 x 1023 Na2CO3 units.
But each formula unit of Na2CO3 contains 2 x 6.023 x 1023 Na + ions and one CO3 -2 irons and
1 x 6.023 x 1023 CO3 -2 ions.
When using the term mole, it is important to specify the formula of the unit to avoid any misunderstanding.
Eg. A mole of oxygen atom (with the formula O) = 6.023 x 1023 Oxygen atoms.
A mole of oxygen molecule (formula O2) = 6.023 x 1023 O2 molecules
(i.e) 2 x 6.023 x 1023 oxygenMolar mass
The molar mass of a substance is the mass of one mole of the substance. The mass and moles can be related by means of the formula.
Mass
Molar Mass= Mole
Eg; Carbon has a molar mass of exactly 12g/mole
Problems
Solved Problems1. What is the mass in grams of a chlorine atom, Cl?
2. What is the mass in grams of a hydrogen chloride, HCl?
Solution
1. 1. The atomic weight of Cl is 35.5 amu,
So the molar mass of Cl is35.5 g/mol.
Dividing 35.5 g (per mole) by 6.023 x 1023 gives the mass of one atom.Mass of a Cl atom = 35.5g
6.023 x 1023
= 5.89 x 10-23 g
2. The molecular weight of HCl = The atomic weight of H + The atomic weight of Cl
= 1 + 35.45 = 36.45 amu
Therefore 1 mole of HCl contains 36.45g of HCl
Mass of an HCl molecule = 36.45g = 6.05 x 10-23g
6.023 x 1023
Solved Problems
1. What is the mass in grams of a calcium atom, Ca?
2. What is mass in grams of an ethanol molecule, C2H5OH?
3. Calculate the mass (in grams) of each of the following species.
a. Na atom b. S atom c. CH3Cl molecule d. Na2SO3 formula unitSolution
1. The atomic weight of Ca is 40.08 amu
Therefore 1 mole of Ca contains 40.08g of Ca
Mass of a Ca molecule = 40.08g
6.023 x 1023
1. The atomic weight of C2H5OH = AW of C2 + AW of H5 + AW of O + AW of H
= 2 x 12 + 5 x 1 + 1 x 16 + 1 x 1
= 46 g
Therefore 1 mole of C2H5OH contains 46g of C2H5OH Mass of a C2H5OH molecule = 46g
6.023 x 1023
= 7.64 x 10-23g
b. S atom
The atomic weight of S = 32.07 amu
1 mole of S contains 32.07g of Na Mass of S molecule = 32.07g
6.023 x 1023
= 5.82 x 10-23gc. The atomic weight of CH3Cl = AW of C + AW of H3 + AW of Cl
= 1 x 12 + 3 x 1 + 1 x 35.45 = 50.45 g
Therefore 1 mole of CH3Cl contains 50.45g of CH3Cl
Mass of a CH3Cl molecule = 50.45g
6.023 x 1023
= 8.38 x 10-23g
x g of ZnI2 = 319.19g of ZnI2 x 0.0654mole ZnI2
= 20.8750gm ZnI2
d. The atomic weight of Na2SO3 = AW of Na2 + AW of S+ AW of O3
= 2 x 22.99 + 1 x 32.07 + 3 x 16
= 126.05 gTherefore 1 mole of Na2SO3 contains 126.05g of Na2SO3 Mass of a Na2SO3 molecule
= 126.05g
6.023 x 1023
= 20.9281 x 10-23g = 2.09 x 10-22gMole Calculations
i. Converting grams of Substances to moles
To illustrate, consider the conversion of grams of ethanol, C2H5OH, to moles of ethanol.
The molar mass of ethanol is 46.1 g/mol,
So, we write 1 mol C2H5OH = 46.1 g of C2 H5OH
Thus, the factor converting grams of ethanol to moles of ethanol is 1mol C2H5OH/46.1g C2H5OH.
To covert moles of ethanol to grams of ethanol, we simply convert the conversion factor (46.1 g C2H5OH/1 mol C2H5OH).
Again, suppose you are going to prepare acetic acid from 10.0g of ethanol, C2H5OH. How many moles of C2H5OH is this?
You convert 10.0g of C2H5OH to moles of C2H5OH by multiplying by the appropriate conversion factor.
Here we have to use for easy calculation
46.1g of C2 H5OH = 1 mole of C2 H5OH and
10.0g of C2 H5OH = x mole of C2 H5OH here x represents the mole of C2 H5OH
Containing 10.0g of C2 H5OHi.e, 46.1g of C2 H5OH = 1 mole of C2 H5OH
10.0g of C2 H5OH x mole of C2 H5OH
x mole of C2 H5OH = 1 mole of C2 H5OH x 10.0g of C2 H5OH
46.1g of C2 H5OH
= 0.2169 moles of C2 H5OH
That means 10.0g C2 H5OH contains 0.2169 moles C2 H5OH
ii. Converting Moles of Substances to Grams
Solved Problems
1. 1. ZnI2, can be prepared by the direct combination of elements. A chemist determines from the amounts of elements that 0.0654 mol ZnI2 can be formed.
Solution
The molar mass of ZnI2 is 65.39 + 2 x 126.90 = 319.19 g/mol Thus
319.19g of ZnI2 = 1mole of ZnI2
x g of ZnI2 = 0.0654mole ZnI2
319.19g of ZnI2 = 1mole of ZnI2
x g of ZnI2 0.0654mole ZnI2
x g of ZnI2 = 319.19g of ZnI2 x 0.0654mole ZnI2
= 20.8750gm ZnI2
Problems
1. H2O2 is a colorless liquid. A concentrated solution of it is used as a source of oxygen for Rocket propellant fuels. Dilute aqueous solutions are used as bleach. Analysis of a solution shows that it contains 0.909 mol H2O2 in 1.00 L of solution. What is the mass of H2O2 in this volume of solution?
The molar mass of H2O2 is 34g/mol
34g of H2O2 = 1 mole of H2O2
0.909g of H2O2 = x mole of H2O2 here x means the mole of H2O2
containing 0.909g of H2O2
Thus 34g of H2O2 = 1 mole of H2O2
x mole of H2O2 0.909g of H2O2
Therefore 0.909g of H2O2 contains 30.906g H2O2
2. Boric acid, H3BO3 is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H3BO3. What is the mass of boric acid in the sample?
The molar mass of H3BO3 is 61.81g/mol
61.81g of H3BO3 = 1 mole of H3BO3
0.543g of H3BO3 = x mole of H3BO3 here x means the mole of H3BO3
containing 0.543g of H3BO3
Thus 61.81g of H3BO3 = 1 mole of H3BO3
x mole of H3BO3 0.543g of H3BO3
Therefore 0.543g of H3BO3 contains 33.56g H3BO3
3. CS2 is a colourless, highly inflammable liquid used in the manufacture of rayon and cellophane. A sample contains 0.0205 mol CS2.Calculate the mass of CS2 in the sample.
The molar mass of CS2 is 76.14g/mol
76.14g of CS2 = 1 mole of CS2
0.0205g of CS2 = x mole of CS2 here x means the mole of CS2
containing 0.0205g of CS2
Thus 76.14g of CS2 = 1 mole of CS2
x mole of CS2 0.0205g of CS2
Therefore 0.0205g of CS2 contains 33.56g CS2
Converting Grams of Substances to Moles
In the preparation of lead (II) chromate PbCrO4, 45.6 g of lead(II)chromate is obtained as a precipitate. How many moles of PbCrO4 is this?
The molar mass of PbCrO4 is 323 g/mol
45.6g of PbCrO4 = x mole of PbCrO4
323 g of PbCrO4 = 1 mole of PbCrO4
Here x means the mole of PbCrO4 containing 45.6g of PbCrO4
Therefore 45.6g of PbCrO4 = x mole of PbCrO4
323 g of PbCrO4 1 mole of PbCrO4
= 0.141mole of PbCrO4
Problems
1. Nitric acid, HNO3 is a colorless, corrosive liquid used in the manufacture of Nitrogen fertilizers and explosives. In an experiment to develop new explosives for mining operations, a 28.5 g sample of HNO3 was poured into a beaker. How many moles of HNO3 are there in this sample of HNO3?
Answer
The molar mass of HNO3 is 63g/mol
28.5g of HNO3 = x mole of HNO3
63g of HNO3 = 1 mole of HNO3
Here x means the mole of HNO3 containing 63g of HNO3
Therefore 28.5g of HNO3 = x mole of HNO3
63.00g of HNO3 1mole of HNO3
= 0.452 mole of HNO3
2. Obtain the moles of substances in the following.
a. 3.43 g of C b. 7.05 g Br2
c. 76g C4 H10 d. 35.4 g Li2 CO3
e. 2.57 g As f. 7.83 g P4
g. 41.4 g N2H4 h. 153 g Al2 (SO4)3
a. 3.43 g of C
The molar mass of C is 12g/mol
3.43g of C = x mole of C
12g of C = 1 mole of C
Here x means the mole of C containing 3.43g of C
Therefore 3.43g of C = x mole of C
12g of C 1 mole of C
= 0.286 mole of C
b. 7.05 g Br2
The molar mass of Br2 is 159.80g/mol
7.05g of Br2 = x mole of Br2
159.80g of Br2 = 1 mole of Br2
Here x means the mole of Br2 containing 7.05g of Br2
Therefore 7.05g of Br2 = x mole of Br2
159.80g of Br2 1 mole of Br2
= 0.0441 mole of Br2
c. 76g C4 H10
The molar mass of C4 H10 is 58g/mol
76g of C4 H10 = x mole of C4 H10
58g of C4 H10 = 1 mole of C4 H10
Here x means the mole of C4 H10 containing 76g of C4 H10
Therefore 76g of C4 H10 = x mole of C4 H10
58g of C4 H10 1 mole of C4 H10
= 1.38 mole of C4 H10
d. 35.4 g Li2 CO3
The molar mass of Li2 CO3 is 57.88g/mol
35.4g of Li2 CO3 = x mole of Li2 CO3
57.88g of Li2 CO3 = 1 mole of Li2 CO3
Here x means the mole of Li2 CO3 containing 35.4g of Li2 CO3
Therefore 35.4g of Li2 CO3 = x mole of Li2 CO3
57.88g of Li2 CO3 1 mole of Li2 CO3
= 0.6116 mole of Li2 CO3
e. 2.57 g As
The molar mass of As is 74.92g/mol
2.57g of As = x mole of As
74.92g of As = 1 mole of As
Here x means the mole of As containing 2.57g of As
Therefore 2.57g of As = x mole of As
74.92g of As 1 mole of As
= 0.0343 mole of As
f. 7.83 g P4
The molar mass of P4 is 123.88g/mol
7.83g of P4 = x mole of P4
123.88g of P4 = 1 mole of P4
Here x means the mole of P4 containing 7.83g of P4
Therefore 7.83g of P4 = x mole of P4
123.88g of P4 1 mole of P4
= 0.0632 mole of P4
g. 41.4 g N2H4
The molar mass of N2H4 is 32.00g/mol
41.4g of N2H4 = x mole of N2H4
32.00g of N2H4 = 1 mole of N2H4
Here x means the mole of N2H4 containing 41.4g of N2H4
Therefore 41.4g of N2H4 = x mole of N2H4
32.00g of N2H4 1 mole of N2H4
= 1.2938 mole of N2H4
h. 153 g Al2 (SO4)3
The molar mass of Al2 (SO4)3 is 342.14g/mol
153g of Al2 (SO4)3 = x mole of Al2 (SO4)3
342.14g of Al2 (SO4)3 = 1 mole of Al2 (SO4)3
Here x means the mole of Al2 (SO4)3 containing 153g of Al2 (SO4)3
Therefore 153g of Al2 (SO4)3 = x mole of Al2 (SO4)3
342.14g of Al2 (SO4)3 1 mole of Al2 (SO4)3
= 0.4472 mole of Al2 (SO4)3
Calculation of the Number of Molecules in a Given Mass
Solved Problem
How many molecules are there in a 3.46 g sample of hydrogen chloride, HCl?
Note: The number of molecules in a sample is related to moles of compound (1 mol HCl = 6.023 x 1023 HCl molecules). Therefore if you first convert grams HCl to moles, then you can convert moles to number of molecules).
Solution
Step 1: convert grams HCl to moles
Molar mass of HCl is 36.45g/mole
That means 36.45g of HCl = 1mole of HCl
Here 3.46g of HCl = x mole of HCl
Therefore 3.46g of HCl = x mole of HCl
36.45g of HCl = 1mole of HCl
3.46g HCl contains 0.0949 mole of HCl
Step 2: convert moles HCl to number of molicules
So that 3.46g of HCl contains 0.0949 x 6.023 x 1023 molecules
= 5.7158 x 1022 molecules
Problems for Practice
1. How many molecules are there in 56mg HCN?
2. Calculate the following
a. Number of molecules in 43g NH3
b. Number of atoms in 32.0 g Br2
c. Number of atoms in 7.46 g Li
Solution
1. How many molecules are there in 56mg HCN?
Molar mass of HCN is 25g/mol
That is 1 mole of HCN contains 25g HCN
1mole of HCN = 25g HCN
x mole of HCN = 0.056g HCN
That is x mole of HCN = 0.056g HCN
1 mole of HCN 25g HCN
x = 0.00224
0.00224mole HCN contains 0.056g HCN
Therefore 0.056g HCN = 0.00224 x 6.023 x 1023 molecules
= 2.24 x 1020 moleculesa. a) Number of molecules in 43g NH3
Molar mass of NH3 is 17g/mol
That is 1 mole of NH3 contains 17g NH3
1mole of NH3 = 17g NH3
x mole of NH3 = 43g NH3 (here x is the mole of 43g NH3)
That is x mole of NH3 = 43g NH3
1 mole of NH3 17g NH3
x = 2.5294
2.5294mole NH3 contains 43g NH3
Therefore 43g NH3 = 2.5294 x 6.023 x 1023 molecules
= 1.52 x 1022 molecules
b. Number of atoms in 32.0 g Br2
Molar mass of Br2 is 159.80g/mol
That is 1 mole of Br2 contains 159.80g NH3
1mole of Br2 = 159.80g NH3
x mole of Br2 = 32.00g Br2 (here x is the mole of 32g Br2)
That is x mole of Br2 = 32.00g Br2
1 mole of Br2 159.80g Br2
x = 0.20025
0.20025mole Br2 contains 32.00g Br2
Therefore 0.20025mole Br2 = 0.20025 x 6.023 x 1023 molecules
= 1.20 x 1023 molecules
c. Number of atoms in 7.46 g Li
Molar mass of Li is 6.94g/mol
That is 1mole of Li contains 6.94g Li
Here we can take
x mole of Li = 7.46g Li here x is x mole of Li contains 7.46g Li
and 1 mole of Li = 6.94g Li
so x = 7.46
6.94
That is 1.07492 moles of Li contains 7.46g of Li
Here 1.07492 moles of Li = 1.07492 x 6.023 x 1023 molecules
= 6.47 x 1023 molecules
Calculation of Empirical Formula from Quantitative Analysis and Percentage composition
Empirical Formula
"An empirical formula (or) simplest formula for a compound is the formula of a substance written with the smallest integer subscripts". For most ionic substances, the empirical formula is the formula of the compound. This is often not the case for molecular substances. For example, the formula of sodium peroxide, an ionic compound of Na+ and O2 2-, is Na2O2. Its empirical formula is NaO. Thus empirical formula tells you the ratio of numbers of atoms in the compound.Solved Problem
A compound has the following composition Mg = 9.76%,S = 13.01%, 0 = 26.01, H2O = 51.22, what is its empirical formula?
Solution
Step I: Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.
Atomic mass of Mg = 24, S = 32, O = 16, H = 1, H2O = 18
Step II: Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.
Magnesium (9.76%) = 9.76 = 0.4067
24
Sulphur (13.01%) = 13.01 = 0.4067
32
Oxygen (26.01%) = 26.01 = 1.6256
16
Water (51.22%) = 51.22 = 2.8456
18
The smallest quotient obtained from above is 0.4067
So
Magnesium (9.76%) = 0.4067 = 1
0.4067
Sulphur (13.01%) = 0.4067 = 1
0.4067
Oxygen (26.01%) = 1.6256 = 4
0.4067 Water (51.22%) = 2.8456 = 7
0.4067
Step II: Multiply the figures, so obtained by a suitable integer of necessary in order to obtain whole number ratio.
Magnesium (9.76%) = 1
Sulphur (13.01%) = 1
Oxygen (26.01%) = 4
Water (51.22%) = 7
Step IV: Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
Hence the empirical formula is Mg SO4.7H2O.
Problems
1. A substance on analysis, gave the following percentage composition, Na = 43.4%, C = 11.3%, 0 =43.3% calculate its empirical formula [Na = 23, C = 12, O = 16].
Answer
Sodium = 43.40 = 1.8869
23.00
Carbon = 11.30 = 0.9417
12.00
Oxygen = 43.30 = 2.7063
16.00
The smallest quotient obtained from above is 0.9417
Sodium = 1.8869 = 2
0.9417
Carbon = 0.9417 = 1
0.9417
Oxygen = 2.7063 = 3
0.9417
Hence the empirical formula is Na2CO3.
2. What is the simplest formula of the compound which has the following percentage composition: Carbon 80%, hydrogen 20%.
Answer
Atomic mass of C = 12 & H = 1
Carbon (80%) = 80.00 = 6.6667
12.00
Hydrogen (20%) = 20.00 = 20.000
1.00
The smallest quotient obtained from above is 6.6667
Carbon (80%) = 6.6667 = 1.0000
6.6667
Hydrogen (20%) = 20.00 = 3.000
6.6667
Hence the empirical formula is CH31. 2 .A compound on analysis gave the following percentage composition: C - 54.54%, H = 9.09%,
0 =36.36%
Atomic mass of C= 12, H=1, O=16Carbon (54.54%) = 54.54 = 4.545
12.00
Hydrogen (9.09%) = 9.09 = 9.09
1
Oxygen (36.36%) = 36.36 = 2.2725
16.00
The smallest quotient from above 2.2725
Carbon = 4.545 = 2
2.2725
Hydrogen = 9.09 = 4
2.2725
Oxygen = 2.2725 = 1
2.2725
Hence Empirical formula is C2H4O